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A hoop of radius r and mass m rotating w...

A hoop of radius r and mass m rotating with an angular velocity `omega_0` is placed on a rough horizontal surface. The initial velocity of the centre of the hoop is zero. What will be the velocity of the centre of the hoop when it ceases ot slip?

A

`(r omega_(0))/(4)`

B

`(r omega_(0))/(3)`

C

`(r omega_(0))/(2)`

D

`r omega_(0)`

Text Solution

Verified by Experts

The correct Answer is:
C
Refer to Fig.

Let `v` be the velocity of the centre of the hoop, when it ceases to ship.
Applying the principle of conservation of angular momentum, we get
`(m r^(2))omega_(0) = mv xx r + mr^(2) omega`
`= mv xx r+ mr^(2) ((v)/(r )) = 2mv r`
`v = ((mr^(2))omega_(0))/(2mr) = (omega_(0)r)/(2)`
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